$3$ Events addition rule probability problem 
The answer given in the porblem is (a) At least $76$%
However, following the formula given for $3$ event addition probability:
$P(A \cup B \cup C ) = P(A) + P(B) + P(C) - P(A \cap B) -  P(A \cap C) - P(B \cap C) + P(A \cap B \cap C).$
I got $(0.25+0.76+0.40)-(0.25$ $\cdot$ $0.76)-(0.25 $$\cdot$ $0.40)- (0.76 $$\cdot$ $0.4) - (0.25 $$\cdot$$ 0.76 $$\cdot$$ 0.40) = 0.759$
Why in this case the answer is at least $76$%? not at most $76$%?
 A: A quick answer to your question: $P(A \cup B \cup C) \ge P(B) = 76\%$ since $B \subseteq A \cup B \cup C$.
Your mistake is due to the incorrect assumption on the independence of events $A,B,C$.  (Hover your mouse onto the tag link and click on "info" to see that the equality $P(A \cap B) = P(A) P(B)$ requires independence of $A$ and $B$.)  However, this is not given in the question, so your calculations for $P(A \cup B \cup C)$ are incorrect.
A: Let's think in terms of words rather than math symbols first. $P(A\cup B\cup C)$ means the probability of event $A$ or $B$ or $C$ or some combination of $A,B,$ and $C$ occurring. If we just had two events $X$ and $Y$, the probability of $X$ or $Y$ happening is bounded below by the probability of $X$ happening. This is because if $X$ happens, then $(X$ or $Y)$ also happens. Similarly the probability of $X$ or $Y$ happening is bounded below by the probability $Y$ happening for the reason that if $Y$ happens, then $(X$ or $Y)$ also happens. For concreteness, you can try thinking about it where $X$ is the event of it raining tomorrow and $Y$ is the event you see two dogs tomorrow. 
So similarly $P(A\cup B \cup C)\geq P(A)$, and also $P(A\cup B \cup C)\geq P(B)$ and also $P(A\cup B \cup C)\geq P(C)$ are all true. However, as $P(B)\geq P(A)$ and $P(B)\geq P(A)$ we have that $P(A\cup B\cup C)\geq P(B)$ captures all three inequalities.
Why is what you wrote false? Well, because you assumed $P(A\cap B)=P(A)P(B)$. This is true when $A$ and $B$ are independent, which we do not know. In fact, since we don't know anything else, the only statement we can make is $P(A\cup B\cup C)\geq P(B)$ and so the correct answer is A.  
